Question

Tangents are drawn to the circle \[{{x}^{2}}+{{y}^{2}}=12\] at the point where it met by the circle \[{{x}^{2}}+{{y}^{2}}-5x+3y-2=0;\] find the point of intersection of these tangents.

Answer 1

Hint:
The given equation of two circles are \[{{x}^{2}}+{{y}^{2}}=12\] (1) and \[{{x}^{2}}+{{y}^{2}}-5x+3y-2=0;\](2). We will find the equation of the common chord of the circle which we can assume as equation (3). Let this line meet the circle 1 (or 2) at A and B. let the tangents to circle 1 at A and B meet at P (a, b), then AB will be the chord of contact of a tangent to circle 1 from P, therefore, we will get the equation of AB (4). Now lines (3) and (4) are identical, now from equating the lines we get the point of intersection of the tangents.

Complete step by step answer:
We know that the given equation of circles is,
\[{{S}_{1}}={{x}^{2}}+{{y}^{2}}=12\]……….. (1)
\[{{S}_{2}}={{x}^{2}}+{{y}^{2}}-5x+3y-2=0;\]…….. (2)
We also know the equation of the common chord of the circles (1) and (2).
\[\begin{align}
  & {{S}_{1}}-{{S}_{2}}=0 \\
 & \Rightarrow {{x}^{2}}+{{y}^{2}}-12-{{x}^{2}}-{{y}^{2}}+5x-3y+2=0 \\
 & \Rightarrow 5x-3y-10=0........(3) \\
\end{align}\].
Let this line meet circle 1 (or 2) at A and B.

Let the tangents to circle 1 at A and B meet at P (a, b), then AB is the chord of contact of the tangent to circle 1 from P.
Therefore, we will get the equation of AB
\[xa+yb-12=0......(4)\]
Now the lines (3) and (4) are identical
\[\begin{align}
  & \Rightarrow \dfrac{a}{b}=\dfrac{b}{-3}=\dfrac{-12}{-10} \\
 & \Rightarrow \dfrac{b}{-3}=\dfrac{6}{5} \\
 & \Rightarrow b=\dfrac{-18}{5} \\
 & \Rightarrow a=6 \\
 & \therefore a=6,b=\dfrac{-18}{5} \\
 & \text{ P=}\left( \text{6,}\dfrac{-18}{5} \right) \\
\end{align}\]
Therefore, the point of intersection of tangent is \[\text{P=}\left( \text{6,}\dfrac{-18}{5} \right)\]

Note:
Another method
Here we are going to use the formula method
we know that the given two equation of circles are
\[{{S}_{1}}={{x}^{2}}+{{y}^{2}}=12\]
\[{{S}_{2}}={{x}^{2}}+{{y}^{2}}-5x+3y-2=0;\]
We know that the chord of intersection is
\[\begin{align}
  & {{S}_{1}}-{{S}_{2}}=0 \\
 & \Rightarrow {{x}^{2}}+{{y}^{2}}-12-{{x}^{2}}-{{y}^{2}}+5x-3y+2=0 \\
 & \Rightarrow 5x-3y-10=0........(3) \\
\end{align}\]
Comparing equation (1) with the below equation, we get
\[\begin{align}
  & lx+my+n=0 \\
 & l=5,m=-3,n=-10 \\
\end{align}\]
We know that
Point of intersection of tangents as given in the concept = \[\left( \dfrac{-{{a}^{2}}l}{n},\dfrac{-{{a}^{2}}m}{n} \right)\]……... (2)
Here \[{{a}^{2}}=12\]
We already found the value of l, m, n. Substituting the value of l, m, n in (2)
Now we get,
\[\begin{align}
  & \Rightarrow \left( \dfrac{-12\times 5}{-10},\dfrac{-12\times -3}{-10} \right) \\
 & \Rightarrow \left( 6,\dfrac{-18}{5} \right) \\
\end{align}\]
Therefore, The point of intersection \[\text{P=}\left( \text{6,}\dfrac{-18}{5} \right)\]